Solution

From the definition we have $$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1.$$ Hence $$7! =5040.$$

Solution

Firstly we can expand the factorials using the definition $$\dfrac{9!}{6!} = \dfrac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}.$$ This can then be simplified by canceling terms on the top and bottom to give $$\dfrac{9!}{6!} = 9 \times 8 \times 7.$$ Therefore, $$\dfrac{9!}{6!} = 504.$$

For more information see simplifying fractions.

Solution

As above start by expanding the factorials $$\dfrac{16!}{13!\cdot 8!} = \dfrac{16 \times 15 \times \dots \times 3 \times 2 \times 1}{(13 \times 12 \times 11 \times \dots \times 3 \times 2 \times 1 )( 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}.$$ Then cancel out any terms that appear on the top and the bottom of the fraction $$\dfrac{16!}{13!\cdot 8!} = \dfrac{16 \times 15 \times 14}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}.$$ This can be canceled further if we factorize the numerator $$\dfrac{16!}{13!\cdot 8!} = \dfrac{8 \times 2 \times 5 \times 3 \times 2 \times 7}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}.$$ Then finally we get $$\begin{align} \dfrac{16!}{13! \cdot 8!} &= \dfrac{1}{4 \times 3},\\ &=\dfrac{1}{12}. \end{align}$$

Solution

$$\begin{align} \dfrac{(n-1)!}{(n+1)!} &= \dfrac{(n-1) \times (n-2) \times \dots \times 2 \times 1}{ (n+1) \times n \times (n-1) \times \dots \times 2 \times 1},\\ &= \dfrac{1}{(n+1) \times n},\\ &= \dfrac{1}{n^2 + n}. \end{align}$$